In a tree that is mono-rooted, lets call the root node as the primary node, the nodes connected to it by edges as the secondary nodes, and those connected to those by edges as the tertiary nodes, etc..
By a complete subtree of a rooted tree $T$, I mean a subtree $t$ of $T$ such that every subtree of $T$ that stems from the root node of $t$ is a subtree of $t$.
Let primary, secondary, tertiary, ..etc subtrees mean those with the respective root nodes. So a primary complete subtree of $T$ can only be $T$ itself.
Now I want to coin a principle of choice over trees such that for every mono-rooted tree $T$, there is a mono-rooted tree $c(T)$ such that each secondary complete subtree of $c(T)$ is some tertiary subtree of $T$, and such that for each secondary complete subtree $t$ of $T$ there is exactly one secondary complete subtree of $t$ that is a secondary complete subtree of $c(T)$
If we add this principle to axioms of $\sf ZF$, would that result in Choice over sets?
Your principle is inconsistent as stated. Suppose $T$ is a tree consisting of a root $r$, a secondary node $n$, and no other nodes. Applying tree choice, we get a tree $c(T)$ such that for each secondary complete subtree $t$ of $T$, there is exactly one secondary complete subtree of $t$ that is a secondary complete subtree of $c(T)$. But $T$ has a unique secondary complete subtree $t$ which consists of just the node $n$, and $t$ has no secondary complete subtrees, so this is a contradiction.
To fix this, you need to assume that every secondary node of $T$ is connected to some tertiary node. This corresponds to the condition in AC that we can get a choice function for any set of nonempty sets, not for any arbitrary set of sets.
Ok, but the modified principle of "tree choice" (TC) - for every tree $T$ in which every secondary node is connected to some tertiary node, there exists a choice tree $c(T)$ as defined in your question - is (rather obviously) equivalent to AC over ZF.
First, assume AC. Let $T$ be a good tree as in the statement of TC. Let $X$ be the set of secondary nodes of $T$. For each node $n\in X$, let $Y_n$ be the set of tertiary nodes connected to $n$. Since $T$ is good, each $Y_n$ is non-empty. Let $f$ be a choice function for $\{Y_n\mid n\in X\}$. For each $n\in X$, let $T_n$ be the tertiary complete subtree of $T$ above $f(n)$. Define $c(T)$ to be $\{r\}\cup \{T_n\mid n\in X\}$, where $r$ is the root (we can take it to be the root of $T$) and we connect each node $f(n)$ directly to the root, so that each tree $T_n$ is now a secondary complete subtree of $c(T)$. Then $c(T)$ is a choice tree for $T$.
Conversely, assume $TC$. Let $X$ be a set of non-empty sets. Define a tree such that $X$ is the root, the elements of $X$ are the secondary nodes, and, for each element $Y\in X$, $\{(Y,y)\mid y\in Y\}$ are the tertiary nodes above $Y$. This is a good tree in the sense of $TC$, because every element $Y\in X$ is a non-empty set, and hence has a tertiary node $(Y,y)$ above it. Let $c(T)$ be a choice tree for $T$. Since $T$ has no nodes above its tertiary nodes, a tertiary subtree of $T$ is just a single node $(Y,y)$. So $c(T)$ consists of a root $r$ and a set of secondary nodes of the form $(Y,y)$. Let $f$ be the set of secondary nodes in $c(T)$. Then $f$ is a choice function for $X$.